ON FUNCTIONAL EQUATIONS RELATED TO ITERATION OF YI’S POLYNOMIALS

ON FUNCTIONAL EQUATIONS RELATED TO ITERATION OF
YI’S POLYNOMIALS
HA HUY KHOAI, VU HOAI AN, AND PHAM NGOC HOA
Abstract. In this paper, we show that the equation P (f1 , ..., fs+1 ) = Q(g1 , ..., gs+1 ),
where P, Q are polynomials in a class of homogeneous polynomials of FermatWaring type, has entire solutions f1 , · · · , fs+1 ; g1 , · · · , gs+1 . Some classes of unique
range sets for linearly non-degenerate holomorphic curves are also obtained.

1. Introduction
We consider the function equation
P (f1 , f2 , · · · , fs+1 ) = Q(g1 , g2 , · · · , gs+1 ),

(1)

where P, Q are polynomials, and fi , gi are entire functions.
Since the paper of Ritt ([11]), the functional equation P (f ) = Q(g) has been
investigated by many authors (see [1], [2], [3], [4], [6], [8], [9], [10]). This equation
related to some other questions, as to find the conditions under which f −1 (X) =
g −1 (Y ), where f, g are polynomials, and X, Y are compact sets; or the question of
finding unique range sets for meromorphic functions.
It is shown in [4] that for a generic pair of polynomials P, Q the equation has no
non-constant solutions in the set of meromorphic functions. Therefore one should
interested to find the solutions for some classes of polynomials. For example, in
[2] and [6] the authors show some classes of pairs P, Q so that the equation has
solutions.
In this paper we consider the equation P (f ) = Q(g), where P, Q are polynomials
of several variables, and of the Fermat-Waring type, obtained by iteration of Yi’s
polynomials. Recall that Yi’s polynomials are defined by
P (z) = az n + bz n−m + c,
where a, b, c are complex numbers. It is well-known that the zeros sets of Yi’s
polynomials, with some additional conditions on m, n, are unique range sets for
meromorphic functions.

We first show a class of such polynomials for which the equation (1) has solutions.
As a consequence we establish some conditions on holomorphic curves f and g,
such that the pull-back divisors νf (X) and νg (Y ) coincide, where X and Y are
hypersurfaces defined by polynomials of the mentioned above class. Some classes of
unique range sets for meromorphic functions are also obtained.
2010 Mathematics Subject Classification. 30D35.
Key words and phrases. Diophantine Equations, unique range sets, holomorphic curves.
1
The work was supported by the National Foundation for Science and Technology Development
(NAFOSTED) and the Vietnam Institute for Advanced study in Mathematics (VIASM).

1


Now let us describe the class of polynomials of Fermat-Waring type considered in
this paper.
Set:
P1 (z1 , z2 ) = a11 z1n + b11 z1n−m z2m + c11 z2n ,
2

P2 (z1 , z2 , z3 ) = a12 P1n (z1 , z2 ) + b12 P1n−m (z1 , z2 )z3nm + c12 z3n ,
...
n
n−m
ni−1 m
ni
Pi (z1 , z2 , ..., zi+1 ) = a1i Pi−1 (z1 , z2 , ..., zi ) + b1i Pi−1
(z1 , z2 , ..., zi )zi+1
+ c1i zi+1
,

...
ns
ns−1 m
n−m
n
,
+ c1s zs+1
(z1 , z2 , ..., zs )zs+1
(z1 , z2 , ..., zs ) + b1s Ps−1
Ps (z1 , z2 , ..., zs+1 ) = a1s Ps−1
Q1 (z1 , z2 ) = a21 z1n + b21 z1n−m z2m + c21 z2n ,
2

Q2 (z1 , z2 , z3 ) = a22 Qn1 (z1 , z2 ) + b22 Q1n−m (z1 , z2 )z3nm + c22 z3n ,
...
n
ni−1 m
ni
Qi (z1 , z2 , ..., zi+1 ) = a2i Qi−1 (z1 , z2 , ..., zi ) + b2i Qn−m
+ c2i zi+1
,
i−1 (z1 , z2 , ..., zi )zi+1
...
ns−1 m
ns
Qs (z1 , z2 , ..., zs+1 ) = a2s Qns−1 (z1 , z2 , ..., zs ) + b2s Qn−m
+ c2s zs+1
,
s−1 (z1 , z2 , ..., zs )zs+1
∗

where a1i = 0, b1i = 0, c1i = 0, a2i = 0, b2i = 0, c2i = 0, i = 1, ..., s; m, n ∈ N , m < n.
Note that Pi (z1 , z2 , ..., zi+1 ), Qi (z1 , z2 , ..., zi+1 ) are polynomials of degree ni .
Set P (z1 , z2 , ..., zs+1 ) = Ps (z1 , z2 , ..., zs+1 ), Q(z1 , z2 , ..., zs+1 ) = Qs (z1 , z2 , ..., zs+1 ).
For entire functions f1 , · · · , fs+1 ; g1 , · · · , gs+1 over C we consider the following equations:
P (f1 , ..., fs+1 ) = Q(g1 , ..., gs+1 ),
(1.1)
P (f1 , ..., fs+1 ) = P (g1 , ..., gs+1 ).
(1.2)
N
Denote by X ( resp. Y ) the hypersurface of Fermat-Waring type in P (C), which
is defined by the equation
PN (z1 , ..., zN +1 ) = 0(resp, QN (z1 , ..., zN +1 ) = 0).

(1.3)

For a holomorphic map f from C to PN (C) we denote by νf (X) the pull-back of the
divisor X in PN (C) by f . We shall prove the following theorems.
Theorem 1.1. Let P, Q be polynomials defined as above, n ≥ 2m + 9, and either
m ≥ 2, (n, m) = 1, or m ≥ 4, and let f1 , · · · , fs+1 ; g1 , · · · , gs+1 be two families
of linearly independent entire functions over C, satisfying the equation (1.1). Then
there exist constants li such that gi = li fi , i = 1, ..., s + 1.
Theorem 1.2. Let f and g be two linearly non-degenerate holomorphic mappings from C to PN (C) with reduced representations f˜ = (f1 , ..., fN +1 ) and g˜ =
(g1 , ..., gN +1 ), respectively. Let X, Y be the Fermat-Waring hypersurfaces defined
as (1.3), and let n ≥ 2m + 9, and either m ≥ 2, (n, m) = 1 or m ≥ 4. If
νf (X) = νg (Y ), then there exist a non-zero entire function h(z) and constants li ,
such that gi = li hfi , i = 1, ..., N + 1.
Corollary 1.3. In the above notations, assume additionally that n ≥ 2m + 9, and
either m ≥ 2, (n, m) = 1 or m ≥ 4. Let f1 , · · · , fs+1 ; g1 , · · · , gs+1 be two families
of linearly independent entire functions over C, satisfying the equation (1.2). Then
there exist constants li such that gi = li fi , i = 1, ..., s + 1.

Next we give an application of Corollary 1.3. Recall that q polynomials of N + 1
variables are said to be in general position if no set of N + 1 polynomials in this
family has common zeros in C N +1 − {0}.
2


Now let given q linear forms of N + 1 variables (q > N ) in general position:
Li = Li (z1 , ..., zN +1 ) = αi,1 z1 + αi,2 z2 + ... + αi,N +1 zN +1 , i = 1, 2, · · · , q.
Consider q homogeneous polynomials:
Y (z1 , z2 ) = az1n + bz1n−m z2m + cz2n ,
T1 (z1 , ..., zN +1 ) = Y (L1 , L2 ), T2 (z1 , ..., zN +1 ) = Y (T1 (z1 , ..., zN +1 ), Ln3 ),
i−1

..., Ti (z1 , ..., zN +1 ) = Y (Ti−1 (z1 , ..., zN +1 ), Lni+1 ),
q−1

n
Tq (z1 , ..., zN +1 ) = Y (Tq−1 (z1 , ..., zN +1 ), Lq+1
),
where a = 0, b = 0, c = 0, q = 2, 3, ... Then Tq (z1 , ..., zN +1 ) is a homogeneous
polynomial of degree nq , and we set T (z1 , z1 , · · · , zN +1 ) = Tq (z1 , ..., zN +1 ).
Denote by Z the hypersurface of Femat-Waring type in P N (C), defined by the
equation
T (z1 , z1 , · · · , zN +1 ) = 0.
Then we have the following
Corollary 1.4. Let f and g be two non-degenerate holomorphic mappings from
C to P N (C). Assume that:
1/ νf (Z) = νg (Z),
2/ n ≥ 2m + 9, and either m ≥ 2, (n, m) = 1 or m ≥ 4.
Then f ≡ g.

Notices that the constants li in the above Theorems and Corollaries can be defined
exactly in terms of the coefficients of polynomials P and Q. Namely, we have
Proposition 1.5. In the notations of Theorem 1.1, the constants li are defined
by
i

i

n
a11 c21 l2n = a21 c11 l1n , b11 c21 l2n−m = b21 c11 l1n−m ; a1i c2i cn1i−1 li+1
= a2i c1i cn2i−1 lin ,
ni−1 (n−m)

ni−1 (n−m)

b1i c2i cn−m
1i−1 li+1
a1s cn1s−1

=

= b2i c1i cn−m
2i−1 li

s
a2s cn2s−1 lsn , b1s cn−m
1s−1

=

, i = 2, ..., s − 1;

ns−1 (n−m) ns−1 m
b2s cn−m
ls+1 , c1s
2s−1 ls

s

n
= c2s ls+1
.

Proposition 1.6. In the notations of Theorem 1.2, the constants li are defined
by:
i

i

n
a11 c21 l1n = a21 c11 l2n , b11 c21 l1n−m = b21 c11 l2n−m , a1i c2i cn1i−1 lin = a2i c1i cn2i−1 li+1
,
ni−1 (n−m)

b1i c2i cn−m
1i−1 li

ni−1 (n−m)

= b2i c1i cn−m

2i−1 li+1
nN −1 m

nN

a1N cn1N −1 lN = a2N cn2N −1 , b1N cn−m
1N −1 lN +1

, i = 2, ..., N − 1;
nN −1 (n−m)

n−m
= b2N c2N
−1 lN

N

n
, c1N = c2N lN
+1 .

Proposition 1.7. In the notations of Corollary 1.3, the constants li satisfy the
following conditions:
i

i

ni−1 (n−m)

n

l1n = l2n , l1n−m = l2n−m ; lin = li+1
, li
ns

ns−1 (n−m)

ls = 1, ls

ns−1 m

ls+1

ni−1 (n−m)

= li+1

, i = 2, ..., s − 1;

ns

= 1, ls+1 = 1.

Remark. Propositions 1.5, 1.6 and 1.7 show that the equations (1.1), (1.2)
have non-constant solutions only in case where the coefficients of P, Q satisfy some
restricted relations. As in [4] this means that the equations have no solutions in
generic case.
3


2. Preliminaries.

Let f be a non-constant holomorphic function on C. For every a ∈ C, expanding
f around a as f = ai (z − a)i ,we define νf (a) = min{i : ai ≡ 0}. For a point ξ ∈ C
we define the function νfξ : C → N by νfξ (a) = νf −ξ (a). Let f be a holomorphic
curve from C to P N (C). For an arbitrary fixed homogeneous coordinate system (z1 :
· · · : zn+1 ) in P N (C) we take a reduced representation of f : f˜ = (f1 : · · · : fN +1 ).
Let H be a hypersurface of P N (C) such that the image of f is not contained in H,
and H is defined by the equation F = 0. For every a ∈ C set
νf (H, a) = νF ◦f˜(a), νf (H) = νf (H, 0).
We assume that the reader is familiar with the notations in the Nevanlinna theory
(see [5 ]) .
The following lemmas were proved in [5].
Lemma 2.1. Let f be a non-constant meromorphic function on C and let a1 , a2 , ..., aq
be distinct points of C ∪ {∞}. Then
q

(q − 2)T (r, f ) ≤

N1 (r,
i=1

1
) + S(r, f ),
f − ai

where S(r, f ) = o(T (r, f )) for all r, except for a set of finite Lebesgue measure.
Lemma 2.2. Let f be a non-constant meromorphic function on C and let a1 , a2 , ..., aq
be distinct points of C ∪ {∞}. Suppose either f − ai has no zeros, or f − ai has
zeros, in which case all the zeros of the functions f − ai have multiplicity at least
mi , i = 1, ..., q. Then
q

1
(1 −
) ≤ 2.
mi
i=1
3. Functional Equations and unique range sets
We first need the following Lemmas:
i
Lemma 3.1. [7 ] Let xd−q
Di (x1 , x2 , ..., xN +1 ),1 ≤ i ≤ N + 1, be homogeneous polyi
nomials of degree d which determine hypersurfaces in general position of P N (C).
Suppose there exists a holomorphic curve f from C to P N (C) with the reduced representation f˜ = (f1 : · · · : fN +1 ), such that its image lies in the curve defined by
N +1

N +1
i
xd−q
Di (x1 , x2 , ..., xN +1 ) = 0, d ≥ N 2 +
i

i=1

qi .
i=1

1
N
Then the polynomials xd−q
D1 (x1 , x2 , ..., xN +1 ), ..., xd−q
DN (x1 , x2 , ..., xN +1 ) are lin1

N
early dependent on the image of f .

Lemma 3.2. Let n, n1 , n2 , ..., nq , q ∈ N∗ , a1 , ..., aq , c ∈ K, c = 0, and q > 2+
Then the functional equation
(f − a1 )n1 (f − a2 )n2 ...(f − aq )nq = cg n
has no non-constant meromorphic solutions (f, g).
4

q
ni
i=1 n .


Proof. Suppose that (f, g) is a non-constant meromorphic solution of the equation:
(f − a1 )n1 (f − a2 )n2 ...(f − aq )nq = cg n .
From this we see that if z0 ∈ C is a zero of f − ai for some 1 ≤ i ≤ q, then z0 is a
zero of g and ni νfai (z0 ) = nνg (z0 ). So
1
ni
1
ni
) ≤ N (r,
) ≤ T (r, f ) + O(1).
f − ai
n
f − ai
n
From this and by Lemma 2.1,
N1 (r,

q

1
(q − 2)T (r, f ) ≤
) + S(r, f ) ≤
N1 (r,
f − ai
i=1

q

ni
1
N (r,
) + S(r, f )
f − ai
i=1 n

q

q
ni
ni
+ O(1) ≤
T (r, f ) + S(r, f ); (q − 2 −
)T (r, f ) ≤ S(r, f ).
i=1 n
i=1 n

Since q > 2 + qi=1 nni , we obtain a contradiction.
Lemma 3.3. Let n, m ∈ N∗ , n ≥ 2m+9, a1 , b1 , c1 , a2 , b2 , c2 ∈ C, a1 , b1 , c1 , a2 , b2 , c2 =
0.
1. Suppose that ff21 is a non-constant meromorphic function, and
f1n + a1 f1n−m f2m + b1 f2n = b2 g2n .

(3.1)

Then there exists a constant h such that g2 = hf2 .
2. Suppose that ff21 and gg12 are non-constant meromorphic functions, and either
m ≥ 2, (m, n) = 1, or m ≥ 4 and
a1 f1n + b1 f1n−m f2m + c1 f2n = a2 g1n + b2 g1n−m g2m + c2 g2n .

(3.2)

Then there exist constants h and l, satisfying a1 = a2 hn , b1 = b2 hn−m lm , c1 = c2 ln ,
such that g1 = hf1 , g2 = lf2 .
Proof.
1. From (3.1) we have
f1n−m (f1m + a1 f2m ) + b1 f2n − b2 g2n = 0.

(3.3)

n
n
m
Note that xn−m
(xm
1
1 + a1 x2 ), b1 x2 , −b2 x3 are the homogeneous polynomials of degree

n in general position. Since n ≥ 2m + 9 and by Lemma 3.1, there exists c1 = 0 such
that c1 b2 g2n = b1 f2n . Therefore g2 = hf2 with b1 = c1 b2 hn .

2. From (3.2 ) we have
a1 f1n + b1 f1n−m f2m + c1 f2n − a2 g1n − b2 g1n−m g2m − c2 g2n = 0,

(3.4)

c1 f2n + f1n−m (a1 f1m + b1 f2m ) − c2 g2n − g1n−m (a2 g1m + b2 g2m ) = 0.

(3.5)

and then
c1 xn1 , xn−m
(a1 xm
2
2

n−m
n
b1 x m
(a2 xm
1 ), −c2 x3 , −x4
4

Note that
+
+ b2 x m
3 ) are the homogeneous polynomials of degree n in general position. Since n ≥ 2m + 9 and by Lemma
3.1, there exist constants C1 , C2 , C3 , (C1 , C2 , C3 ) = (0, 0, 0), such that

C1 c1 f2n + C2 f1n−m (a1 f1m + b1 f2m ) + C3 c2 g2n = 0.
We consider the following possible subcases:
5

(3.6)


Subcase 1: C3 = 0. Then from (3.6) we have
C1 c1 f2n + C2 f1n−m (a1 f1m + b1 f2m ) = 0.
Since f2 is a non-zero entire function, we have C2 = 0. If C1 = 0, then ff21 is a
constant, a contradiction. So C1 , C2 = 0. Then ff12 is a constant, a contradiction. So
C3 = 0.
Subcase 2: C2 = 0. Then from (3.6) we have C1 c1 f2n + C3 c2 g2n = 0. Because f2 , g2
are non-zero entire functions, we have C1 = 0, C3 = 0. From this and (3.5) it follows
C1 c1 n g2
= l, l = 0, and
that g2n = − C
f ,
3 c2 2
f2
C1 n
c1 1 +
f + f1n−m (a1 f1m + b1 f2m ) − g1n−m (a2 g1m + b2 g2m ) = 0.
C3 2
C1 n−m
−a2 g1n + f1n−m (a1 f1m + b1 f2m ) + c1 (1 +
− b2 lm g1n−m f2m = 0.
)f
(3.7)
C3 2

C1
= 0. Then, from the similarity of (3.7) and (3.5), by a similar
Suppose that 1 +
C3
argument as in (3.9), there exist constants C1 , C2 , (C1 , C2 ) = (0, 0), such that
C2 a2 g1n + C1 f1n−m (a1 f1m + b1 f2m ) = 0.
Since g1 is a non-zero entire function and
C1 = 0, C2 = 0. We have

f1
f2

(3.8).

is not a constant, by (3.8) we obtain

f1 n
f1 n−m
g1 n
+ C 1 b1
= −C2 a2
.
f2
f2
f2
f1 n−m f1 m b1
g1 n
C 1 a1
.
(3.9)

( ) +
= −C2 a2
f2
f2
a1
f2
Note that the equation z m + ab11 = 0 has m distinct roots d1 , d2 , ..., dm . Set f =
f1
, g = fg12 . Consequently, by (3.9) we have
f2
C1 f1n−m (a1 f1m + b1 f2m ) = −C2 a2 g1n , C1 a1

f n−m (f − d1 )...(f − dm ) = Cg n , C = 0.

(3.10)

Since ff21 is not a constant, neither is fg12 . If (m, n) = 1, then from (3.10) we see
that the multiplicity of each zero of f and f − di is a multiple of n. By n ≥ 2m +
9, m ≥ 2 and Lemma 2.2 we conclude that the equation (3.10) has no non-constant
+
meromorphic solutions. If m ≥ 4, then from (3.10) and we have m + 1 > 2 + n−m
n
m 1
i=1 n . Then applying Lemma 3.2 to (3.10) with q = m + 1, n = n, n1 = n − m, n2 =
C1
1 = n3 = ... = nm , we have a contradiction. So 1 +
= 0. Therefore c2 g2n = c1 f2n ,
C3
and g2 = lf2 with c1 = c2 ln .
Subcase 3. C1 = 0. From (3.6) we have C2 f1n−m (a1 f1m + b1 f2m ) + C3 c2 g2n = 0. Then,

from the similarity of this equation and (3.8), by a similar argument as in (3.8) we
have a contradiction.
Subcase 4. C1 = 0, C2 = 0, C3 = 0.
By a similar argument as in (3.7) we obtain a contradiction. So c2 g2n = c1 f2n , and
g2 = lf2 with c1 = c2 ln .
6


Moreover from (3.2) we have
c1 f2n (
where f =

a1 n b1 n−m
a2
b2
f + f
+ 1) = c2 g2n ( g n + g n−m + 1),
c1
c1
c2
c2

f1
g1
a1
b1
a2
b2
, g = . Set
= a3 , = b 3 ,

= a4 , = b4 . Since c2 g2n = c1 f2n ,
f2
g2
b1
c1
c2
c2
a3 f n + b3 f n−m = a4 g n + b4 g n−m .

Set h1 =

g
. From this we obtain
f

g n
g n−m
,
a3 f m + b 3 = a4 ( ) f m + b 4 ( )
, a3 f m + b3 = a4 hn1 f m + b4 hn−m
1
f
f
a3
)
1
a4
f m (a3 − a4 hn1 ) = b4 h1n−m − b3 , −
= ( )m .
(3.11)

b3
f
n−m
b4 (h1
− )
b4
Assume that h1 is not a constant. Consider the following possible cases:
a3
b3
Case 1. m ≥ 2, (m, n) = 1. If hn1 −
and h1n−m − have no common zeros, then
a4
b4
a3
n
have multipcities ≥ m. Then
all zeros of h1 −
a4
a4 (hn1 −

N1 (r,

1

1
1
) ≤ N (r,
a
a3 ).
3

m
hn1 −
hn1 −
a4
a4

By Lemma 2.1 we obtain
1
1
) + N1 (r,
a3 ) + S(r, f ),
n
h1
hn1 −
a4
n
n
1
nT (r, h1 ) ≤ 2T (r, h1 ) + N (r,
)
+
S(r,
f
)
≤
(2
+
)T (r, h1 ) + S(r, f )
a
3

m
m
h1 −
a4
n
(n − 2 − )T (r, h1 ) ≤ S(r, f ),
m
which leads to n(m − 1) ≤ 2m, a contradiction to the condition n ≥ 2m + 9. If
a3
b3
a3
hn1 −
and hn−m
− have common zeros, then there exists z such that hn1 (z) = ,
1
a4
b4
a4
b
3
hn−m
(z) = . Set a = h1 (z). From (3.11) we get
1
b4
T (r, hn1 ) ≤ N1 (r, hn1 ) + N1 (r,

a4 (hn1 − an )
1
−
= ( )m , −

n−m
n−m
b4 (h1
−a
)
f

h1 n
) − 1)
1
a4 am (hn2 − 1)
1
a
= ( )m , −
= ( )m ,
n−m
h
f
b4 (h2
− 1)
f
1
b4 an−m (( )n−m − 1)
a
a4 an ((

7


h1

where h2 = . Since (m, n) = 1, the equations z n − 1 = 0 and z n−m − 1 = 0 have
a
different roots except for z = 1. Let ri , i = 1, ..., 2n − m − 2 be all the roots of them.
Then all zeros of h2 − ri have multipcities ≥ m. Therefore, by Lemma 2.2 we obtain
1
(1 − )(2n − m − 2) ≤ 2, 2n(m − 1) ≤ m2 + 3m − 2,
m
which contradicts n ≥ 2m + 9. Thus h1 is a constant.
a3
= 0 has n simple zeros, equation
Case 2. m ≥ 4. Note that equation z n −
a4
b3
a3
b3
z n−m −
= 0 has n − m simple zeros. Then z n −
= 0, z n−m −
= 0 have
b4
a4
b4
a3
= 0 has at
at most n − m common simple zeros. Therefore the equation z n −
a4
b3
least m distinct roots, which are not roots of z n−m −
= 0. Let r1 , r2 , ..., rm be
b4

all these roots. Then all the simple zeros of h1 − rj , j = 1, ..., m, have multiplicities
1
≥ m. By Lemma 2.2 we have m(1 − ) ≤ 2. Therefore m ≤ 3. From m ≥ 4,
m
we obtain a contradiction. Thus h1 is a constant. By g = h1 f, g2 = lf2 we have
g1 = hf1 . From (3.2) and since ff21 and gg12 are non-constant functions we obtain
a1 = a2 hn , b1 = b2 hn−m lm , c1 = c2 ln . Lemma 3.3 is proved.
Now we use the above Lemmas to prove the main result of the paper.
Proof of Theorem 1 and Proposition 1.5
Set Pi (f˜) = Pi (f1 , ..., fi+1 ), Pi (˜
g ) = Pi (g1 , ..., gi+1 ), i = 1, ..., s. We first prove
˜
Pi (f ) ≡ 0, i = 1, 2, ..., s, by induction on i. With i = 1 assume that
P1 (f˜) = a11 f1n + b11 f1n−m f2m + c11 f2n ≡ 0.
f1
is a constant, and we have a contradiction to the linearly independence
f2
of f1 , ..., fs+1 . With i = 2, assume that
Therefore,

2
P2 (f˜) = a12 P1n (f˜) + b12 P1n−m (f˜)f3nm + c12 f3n ≡ 0.

P1 (f˜)
is a constant. Hence
Since P1 (f˜) ≡ 0, f3n ≡ 0 we see that
f3n
a11 f1n + b11 f1n−m f2m + c11 f2n − Af3n ≡ 0.
f2
is a constant, and we have a contradiction

f3
to the linearly independence of f1 , ..., fs+1 .
Now we consider Pi (f˜) ≡ 0. Then
Applying Lemma 3.3 we obtain that

n
n−m ˜ ni−1 m
ni
a1i Pi−1
(f˜) + b1i Pi−1
(f )fi+1 + c1i fi+1
≡ 0.

(3.12)

Applying the induction hypothesis and by a similar argument as above we have a
contradiction.
So Pi (f˜) ≡ 0, i = 1, 2, ..., s. Similarly, Qi (f˜) ≡ 0, i = 1, 2, ..., s.
8


Next we consider
Ps (f˜) = Qs (˜
g ).

(3.13)

We will show that gi = li fi , i = 1, ..., s + 1, where li satisfy following conditions:
i

i

n
a11 c21 l2n = a21 c11 l1n , b11 c21 l2n−m = b21 c11 l1n−m ; a1i c2i cn1i−1 li+1
= a2i c1i cn2i−1 lin ,
ni−1 (n−m)

b1i c2i cn−m
1i−1 li+1
a1s cn1s−1

=

ni−1 (n−m)

= b2i c1i cn−m
2i−1 li

s
a2s cn2s−1 lsn , b1s cn−m
1s−1

=

, i = 2, ..., s − 1;

ns−1 (n−m) ns−1 m
b2s cn−m
ls+1 , c1s
2s−1 ls

s

n
= c2s ls+1
.

From (3.13) we have
ns
ns−1 m
n−m
ns
n−m ˜ ns−1 m
n
.
+ c1s gs+1
(˜
g )gs+1
g ) + b2s Qs−1
= a2s Qns−1 (˜
(f )fs+1 + c1s fs+1
(f˜) + b1s Ps−1
a1s Ps−1
(3.14)
Applying Lemma 3.3 for (3.14) we have
s−1 m

n−m n
Qs−1 (˜
g ) = hs−1 Ps−1 (f˜), gs+1 = ls+1 fs+1 ; a1s = a2s hns−1 , b1s = b2s hs−1

ls+1
s

n
c1s = c2s ls+1
.

,

(3.15)

From (3.15) we obtain
s−1
n
n−m ˜ ns−2 m
hs−1 a1s−1 Ps−2
(f˜) + hs−1 b1s−1 Ps−2
(f )fs
+ hs−1 c1s−1 fsn =
s−2 m

n−m
a2s−1 Qns−2 (˜
g ) + b2s−1 Qs−2
(˜
g )gsn

s−1

+ c2s−1 gsn

.

(3.16)

Applying Lemma 3.3 for (3.16) we have
Qs−2 (˜
g ) = hs−2 Ps−2 (f˜), gs = ls fs ; hs−1 a1s−1 = a2s−1 hns−2 ,
s−2

s−1

n
m
hs−1 b1s−1 = b2s−1 hn−m
, hs−1 c1s−1 = c2s−1 lsn .
s−2 ls
Since and (3.15) we give
c2s−1 ns−1
c2s−1 ns−1 n
c2s−1 ns−1 n−m ns−1 m
hs−1 =
ls , a1s = a2s (
ls ) , b1s = b2s (
l
)
ls+1 ,
c1s−1
c1s−1
c1s−1 s

s

s−1 (n−m)

n−m n
a1s cn1s−1 = a2s cn2s−1 lsn , b1s cn−m
1s−1 = b2s c2s−1 ls

s−1 m

n
ls+1

s

n
, c1s = c2s ls+1
.

(3.17)

Now we consider following
Qi (˜
g ) = hi Pi (f˜), i = 2, ..., s − 1;
i−1 m

n
+ hi c1i fi+1
=

i−1 m

n
+ c2i gi+1
.

n
n−m ˜ n
hi a1i Pi−1
(f˜) + hi b1i Pi−1
(f )fi+1
n−m
n
a2i Qni−1 (˜
g ) + b2i Qi−1
(˜
g )gi+1

i

i

(3.18)

Applying Lemma 3.3 for (3.18) we give
Qi−1 (˜
g ) = hi−1 Pi−1 (f˜), gi+1 = li+1 fi+1 ; hi a1i = a2i hni−1 ,
c2i ni
ni
ni−1 m

l .
hi b1i = b2i hn−m
, hi c1i = c2i li+1
; hi =
i−1 li+1
c1i i+1
From Qi−1 (˜
g ) = hi−1 Pi−1 (f˜) we give
i−2 m

n
n−m ˜ n
hi−1 a1i−1 Pi−2
(f˜) + hi−1 b1i−1 Pi−2
(f )fi

9

i−1

+ hi−1 c1i−1 fin

(3.19)

=


i−2 m

n−m

a2i−1 Qni−2 (˜
g ) + b2i−1 Qi−2
(˜
g )gin

i−1

+ c2i−1 gin

.

(3.20)

Applying Lemma 3.3 for (3.20) we obtain
Qi−2 (˜
g ) = hi−2 Pi−2 (f˜), gi = li fi ; hi−1 a1i−1 = a2i−1 hni−2 ,
c2i−1 ni−1
i−1
ni−2 m
, hi−1 c1i−1 = c2i−1 lin ; hi−1 =
.
l
hi−1 b1i−1 = b2i−1 hn−m
i−2 li
c1i−1 i
From this and (3.19) we obtain
c2i−1 ni−1
c2i ni c2i ni
c2i−1 ni−1 n
hi−1 =

li , hi =
li+1 ,
li+1 a1i = a2i (
l
) ,
c1i−1
c1i
c1i
c1i−1 i
c2i ni
c2i−1 ni−1 n−m ni−1 m
i
ni
li+1 b1i = b2s (
li )
li+1 , a1i c2i cn1i−1 li+1
= a2i c1i cn2i−1 lin ;
c1i
c1i−1
ni−1 (n−m)

b1i c2i cn−m
1i−1 li+1

ni−1 (n−m)

n−m
li
= b2i c1i c2i−1

.

(3.21)

Now we consider the following equations:
Q1 (˜
g ) = h1 P1 (f˜),
h1 a11 f1n + h1 b11 f1n−m f2m + h1 c11 f2n = a21 g1n + b22 g1n−m g2m + c22 g2n .
Applying Lemma 3.3 for (3.22) we have

(3.22)

g1 = l1 f1 , g2 = l2 f2 , h1 a11 = a21 l1n , h1 b11 = b21 l1n−m l2m , h1 c11 = c21 l2n .
Form this it follows
c21 n c21
c21
h1 =
l2 ,
a11 l2n = a21 l1n ,
b11 l2n = b21 l1n−m l2m .
c11
c11
c11
Therefore
a11 c21 l2n = a21 c11 l1n , b11 c21 l2n−m = b21 c11 l1n−m
(3.23)
By (3.17), (3.21), (3.23), Theorem 1.1 is proved.
Now we are going to complete the proof of Theorem 2.
Proof of Theorem 2 and Propositions 1.6.
Let f˜ = (f1 , ..., fN +1 ) and g˜ = (g1 , ..., gN +1 ) be reduced representations of f and

g, respectively.
Since νf (X) = νg (Y ), it is easy to see that there exists a non-zero entire function
N
h(z) such that such that Pq (f˜) = hQq (˜
g ). Take b such that bn = h and set
ϕ = (bg1 : · · · : bgN +1 ). Then ϕ is a reduced representation of g and PN (f˜) = QN (ϕ).
By Theorem 2 then follows from Theorem 1.
Corollary 1.3 and Proposition 1.7 are consequences of Theorem 1.1 and Proposition 1.5.
Now we are going to complete the proof of Corollary 1.4.
Proof of Corolary 1.4.
Let f˜ = (f1 , ..., fN +1 ) and g˜ = (g1 , ..., gN +1 ) be reduced representations of f and
g, respectively.
Since νf (X) = νg (X), it is easy to see that there exists an non-zero entire function
N
h such that Pq (f˜) = hPq (˜
g ). Take b such that bn = h and set Gi = bgi , i =
˜ = (G1 : · · · : GN +1 ). Then G
˜ is a reduced representation of g and
1, ..., N + 1, G
˜
Pq (f˜) = Pq (G).
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˜ = Li (G1 , · · · , GN +1 ), i = 1, ..., q, Pi (f˜) =
Set Li (f˜) = Li (f1 , · · · , fN +1 ), Li (G)
˜
Pi (f1 , ..., fN +1 ), Pi (G) = Pi (G1 , ..., GN +1 ), i = 1, ..., q. By Corollary 1.3 we obtain
˜ = ci Li (f˜), ci = 0, i = 1, ..., q + 1. Since Li , i = 1, ..., N + 1, are linearly
Li (G)

independent and L1 , ..., LN +1 , Lj , j ∈ {N + 2, ..., q + 1} are linearly dependent, we
get
Lj = b1j L1 + b2j L2 + ... + bN +1j LN +1 , bkj = 0, k = 1, ..., N + 1, j = N + 2, ..., q + 1;
Lj (f˜) = b1j L1 (f˜) + b2j L2 (f˜) + ... + bN +1j LN +1 (f˜), j = N + 2, ..., q + 1;
˜ = b1j L1 (G)
˜ + b2j L2 (G)
˜ + ... + bN +1j LN +1 (G),
˜ j = N + 2, ..., q + 1.
Lj (G)
˜ = ci Li (f˜), ci = 0, i = 1, 2, ..., N + 1; Lj (G)
˜ = cj Lj (f˜), we
From this and Li (G)
obtain
˜ = c1 b1j L1 (f˜) + c2 b2j L2 (f˜) + ... + cN +1 bN +1j LN +1 (f˜);
Lj (G)
c1 b1j L1 (f˜) + c2 b2j L2 (f˜) + ... + cN +1 bN +1j LN +1 (f˜) =
cj b1j L1 (f˜) + cj b2j L2 (f˜) + ... + cj bN +1j LN +1 (f˜), j = N + 2, ..., q + 1.
By the linearly independence of f1 , ..., fN +1 we obtain cj = c1 = cj = c2 = ... =
˜ = cLj (f˜), j =
cN +1 , j = N + 2, ..., q + 1. Set c = ci , i = 1, ..., q + 1. Then Lj (G)
1, ..., q + 1. Then Gi = cfi , i = 1, ..., N + 1. So f ≡ g.
References
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2011, pp.253-262.
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Appl. 351 (2009), 350-359.
[3] Ha Huy Khoai and Vu Hoai An and Le Quang Ninh Uniqueness Theorems for Holomorphic
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Institute of Mathematics,VAST, and Thang Long University, Hanoi, Viet Nam
E-mail address:
Hai Duong College, Hai Duong, Viet Nam
E-mail address:
E-mail address:
Hai Duong Pedagogical College, Hai Duong, Viet Nam

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